# How do you find the vertex and intercepts for y=x^2 + 4x + 2?

Complete the square. $y = {\left(x + \frac{4}{2}\right)}^{2} - {2}^{2} + 2 = {\left(x + 2\right)}^{2} - 2$. So the vertex is at $\left(- 2 , - 2\right)$
The minimum possible value of ${\left(x + 2\right)}^{2}$ is zero and will occur when $x = - 2$. To complete the square, add half the coefficient of $x$ to the ${x}^{2}$ term, square the sum, and take away the square of the number you added. thus returning the expression to its original value.
The $y$-intercept is $\left(0 , 2\right)$. The $x$-co-ordinates of the $x$-intercepts are the roots of the equation ${x}^{2} + 4 x + 2 = 0$, namely $- 2 \pm \sqrt{2}$.